Strength of Materials Prof. M. S. Sivakumar

Indian Institute of Technology Madras

Axial Deformations

Introduction

Free body diagram - Revisited

Normal, shear and bearing stress

Stress on inclined planes under axial loading

Strain

Mechanical properties of materials

True stress and true strain

Poissons ratio

Elasticity and Plasticity

Creep and fatigue

Deformation in axially loaded members

Statically indeterminate problems

Thermal effect

Design considerations

Strain energy

Impact loading

Strength of Materials Prof. M. S. Sivakumar

Indian Institute of Technology Madras

1.1 Introduction

An important aspect of the analysis and design of structures relates to the deformations

caused by the loads applied to a structure. Clearly it is important to avoid deformations so

large that they may prevent the structure from fulfilling the purpose for which it is intended.

But the analysis of deformations may also help us in the determination of stresses. It is not

always possible to determine the forces in the members of a structure by applying only the

principle of statics. This is because statics is based on the assumption of undeformable,

rigid structures. By considering engineering structures as deformable and analyzing the

deformations in their various members, it will be possible to compute forces which are

statically indeterminate. Also the distribution of stresses in a given member is

indeterminate, even when the force in that member is known. To determine the actual

distribution of stresses within a member, it is necessary to analyze the deformations which

take place in that member. This chapter deals with the deformations of a structural

member such as a rod, bar or a plate under axial loading.

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Strength of Materials Prof. M. S. Sivakumar

Indian Institute of Technology Madras

1.2 Free body diagram - Revisited

The first step towards solving an engineering problem is drawing the free body diagram of

the element/structure considered.

Removing an existing force or including a wrong force on the free body will badly affect the

equilibrium conditions, and hence, the analysis.

In view of this, some important points in drawing the free body diagram are discussed

below.

Figure 1.1

At the beginning, a clear decision is to be made by the analyst on the choice of the body to

be considered for free body diagram.

Then that body is detached from all of its surrounding members including ground and only

their forces on the free body are represented.

The weight of the body and other external body forces like centrifugal, inertia, etc., should

also be included in the diagram and they are assumed to act at the centre of gravity of the

body.

When a structure involving many elements is considered for free body diagram, the forces

acting in between the elements should not be brought into the diagram.

The known forces acting on the body should be represented with proper magnitude and

direction.

If the direction of unknown forces like reactions can be decided, they should be indicated

clearly in the diagram.

Strength of Materials Prof. M. S. Sivakumar

Indian Institute of Technology Madras

After completing free body diagram, equilibrium equations from statics in terms of forces

and moments are applied and solved for the unknowns.

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Strength of Materials Prof. M. S. Sivakumar

Indian Institute of Technology Madras

1.3 Normal, shear and bearing stress

1.3.1 Normal Stress:

Figure 1.2

When a structural member is under load, predicting its ability to withstand that load is not

possible merely from the reaction force in the member.

It depends upon the internal force, cross sectional area of the element and its material

properties.

Thus, a quantity that gives the ratio of the internal force to the cross sectional area will

define the ability of the material in with standing the loads in a better way.

That quantity, i.e., the intensity of force distributed over the given area or simply the force

per unit area is called the stress.

P

A

σ

=

1.1

In SI units, force is expressed in newtons (N) and area in square meters. Consequently,

the stress has units of newtons per square meter (N/m

2

) or Pascals (Pa).

In figure 1.2, the stresses are acting normal to the section XX that is perpendicular to the

axis of the bar. These stresses are called normal stresses.

The stress defined in equation 1.1 is obtained by dividing the force by the cross sectional

area and hence it represents the average value of the stress over the entire cross section.

Strength of Materials Prof. M. S. Sivakumar

Indian Institute of Technology Madras

Figure 1.3

Consider a small area

∆

A on the cross section with the force acting on it

∆

F as shown in

figure 1.3. Let the area contain a point C.

Now, the stress at the point C can be defined as,

A 0

F

lim

A

∆ →

∆

σ =

∆

1.2

The average stress values obtained using equation 1.1 and the stress value at a point from

equation 1.2 may not be the same for all cross sections and for all loading conditions.

1.3.2 Saint - Venant's Principle:

Figure 1.4

Strength of Materials Prof. M. S. Sivakumar

Indian Institute of Technology Madras

Consider a slender bar with point loads at its ends as shown in figure 1.4.

The normal stress distribution across sections located at distances b/4 and b from one and

of the bar is represented in the figure.

It is found from figure 1.4 that the stress varies appreciably across the cross section in the

immediate vicinity of the application of loads.

The points very near the application of the loads experience a larger stress value whereas,

the points far away from it on the same section has lower stress value.

The variation of stress across the cross section is negligible when the section considered

is far away, about equal to the width of the bar, from the application of point loads.

Thus, except in the immediate vicinity of the points where the load is applied, the stress

distribution may be assumed to be uniform and is independent of the mode of application

of loads. This principle is called Saint-Venant's principle.

1.3.3 Shear Stress:

The stresses acting perpendicular to the surfaces considered are normal stresses and

were discussed in the preceding section.

Now consider a bolted connection in which two plates are connected by a bolt with cross

section A as shown in figure 1.5.

Figure 1.5

The tensile loads applied on the plates will tend to shear the bolt at the section AA.

Hence, it can be easily concluded from the free body diagram of the bolt that the internal

resistance force V must act in the plane of the section AA and it should be equal to the

external load P.

These internal forces are called shear forces and when they are divided by the

corresponding section area, we obtain the shear stress on that section.

Strength of Materials Prof. M. S. Sivakumar

Indian Institute of Technology Madras

V

A

τ

=

1.3

Equation 1.3 defines the average value of the shear stress on the cross section and the

distribution of them over the area is not uniform.

In general, the shear stress is found to be maximum at the centre and zero at certain

locations on the edge. This will be dealt in detail in shear stresses in beams (module 6).

In figure 1.5, the bolt experiences shear stresses on a single plane in its body and hence it

is said to be under single shear.

Figure 1.6

In figure 1.6, the bolt experiences shear on two sections AA and BB. Hence, the bolt is

said to be under double shear and the shear stress on each section is

V P

A 2A

τ = =

1.4

Assuming that the same bolt is used in the assembly as shown in figure 1.5 and 1.6 and

the same load P is applied on the plates, we can conclude that the shear stress is reduced

by half in double shear when compared to a single shear.

Shear stresses are generally found in bolts, pins and rivets that are used to connect

various structural members and machine components.

1.3.4 Bearing Stress:

In the bolted connection in figure 1.5, a highly irregular pressure gets developed on the

contact surface between the bolt and the plates.

The average intensity of this pressure can be found out by dividing the load P by the

projected area of the contact surface. This is referred to as the bearing stress.

Strength of Materials Prof. M. S. Sivakumar

Indian Institute of Technology Madras

Figure 1.7

The projected area of the contact surface is calculated as the product of the diameter of

the bolt and the thickness of the plate.

Bearing stress,

b

P P

A t d

σ = =

×

1.5

Example 1:

Figure 1.8

Strength of Materials Prof. M. S. Sivakumar

Indian Institute of Technology Madras

A rod R is used to hold a sign board with an axial load 50 kN as shown in figure 1.8. The

end of the rod is 50 mm wide and has a circular hole for fixing the pin which is 20 mm

diameter. The load from the rod R is transferred to the base plate C through a bracket B

that is 22mm wide and has a circular hole for the pin. The base plate is 10 mm thick and it

is connected to the bracket by welding. The base plate C is fixed on to a structure by four

bolts of each 12 mm diameter. Find the shear stress and bearing stress in the pin and in

the bolts.

Solution:

Shear stress in the pin =

(

)

( )

3

pin

2

50 10/2

V P

79.6 MPa

A 2A

0.02/4

×

τ = = = =

π

Force acting on the base plate = Pcos

θ

= 50 cos30

0

= 43.3 kN

Shear stress in the bolt,

(

)

( )

3

bolt

2

43.3 10/4

P

4A

0.012/4

×

τ = =

π

95.7 MPa=

Bearing stress between pin and rod,

(

)

( ) (

)

3

b

50 10

P

b

d 0.05 0.02

×

σ = =

× ×

50 MPa=

Bearing stress between pin and bracket =

(

)

( ) (

)

3

b

50 10/2

P/2

b

d 0.022 0.02

×

σ = =

× ×

= 56.8 MPa

Bearing stress between plate and bolts =

(

)

( ) (

)

3

b

43.3 10/4

P/4

t d 0.01 0.012

×

σ = =

× ×

90.2 MPa=

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Strength of Materials Prof. M. S. Sivakumar

Indian Institute of Technology Madras

1.4 Stress on inclined planes under axial loading:

When a body is under an axial load, the plane normal to the axis contains only the normal

stress as discussed in section 1.3.1.

However, if we consider an oblique plane that forms an angle

?

with normal plane, it

consists shear stress in addition to normal stress.

Consider such an oblique plane in a bar. The resultant force P acting on that plane will

keep the bar in equilibrium against the external load P' as shown in figure 1.9.

Figure 1.9

The resultant force P on the oblique plane can be resolved into two components F

n

and F

s

that are acting normal and tangent to that plane, respectively.

If A is the area of cross section of the bar, A/cos

?

is the area of the oblique plane. Normal

and shear stresses acting on that plane can be obtained as follows.

F

n

= Pcos

θ

F

s

= -Psin

θ

(Assuming shear causing clockwise rotation negative).

2

Pcos P

cos

A/cos A

θ

σ = = θ

θ

1.6

Psin P

sin cos

A/cos A

θ

τ = − = − θ θ

θ

1.7

Equations 1.6 and 1.7 define the normal and shear stress values on an inclined plane that

makes an angle

θ

with the vertical plane on which the axial load acts.

From above equations, it is understandable that the normal stress reaches its maximum

when

θ

= 0

o

and becomes zero when

θ

= 90

o.

Strength of Materials Prof. M. S. Sivakumar

Indian Institute of Technology Madras

But, the shear stress assumes zero value at

θ

= 0

o

and

θ

= 90

o

and reaches its maximum

when

θ

= 45

o

.

The magnitude of maximum shear stress occurring at

θ

= 45

o

plane is half of the maximum

normal stress that occurs at

θ

= 0

o

for a material under a uniaxial loading.

max

max

P

2A 2

σ

τ = =

1.8

Now consider a cubic element A in the rod which is represented in two dimension as

shown in figure 1.10 such that one of its sides makes an angle

?

with the vertical plane.

Figure 1.10

To determine the stresses acting on the plane mn, equations 1.6 and 1.7 are used as such

and to knows the stresses on plane om,

θ

is replaced by

θ

+ 90

o

.

Maximum shear stress occurs on both om and mn planes with equal magnitude and

opposite signs, when mn forms 45

o

angle with vertical plane.

Example 2:

A prismatic bar of sides 40 mm x 30 mm is axially loaded with a compressive force of 80

kN. Determine the stresses acting on an element which makes 30

0

inclination with the

vertical plane. Also find the maximum shear stress value in the bar.

Strength of Materials Prof. M. S. Sivakumar

Indian Institute of Technology Madras

Figure 1.11

Solution:

Area of the cross section, A = 40 x 30 x 10

-6

= 1.2 x 10

-3

m

2

Normal stress on 30

0

inclined plane,

2

P

cos

A

σ

= θ

3

2 o

3

80 10

cos 30 50 MPa

1.2 10

−

− ×

= × = −

×

Shear stress on 300 plane,

3

0 0

3

P 80 10

sin cos sin30 cos30

A

1.2 10

−

− ×

τ = θ θ = × ×

×

= 28.9 MPa [Counter clockwise]

Normal stress on 120

0

plane,

3

2 0

3

80 10

cos 120 16.67 MPa

1.2 10

−

− ×

σ = = −

×

Shear stress on 120

0

plane,

3

0 0

3

80 10

sin120 cos120 28.9 MPa

1.2 10

−

×

τ = × × = −

×

[Clock wise]

−

×

τ = =

× ×

±

3

max

3

P 80 10

Maximum shear stress in the bar,

2A

2 1.2 10

= 33.3 MPa

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Strength of Materials Prof. M. S. Sivakumar

Indian Institute of Technology Madras

1.5 Strain

The

structural member and machine components undergo deformation as they are brought

under loads.

To ensure that the deformation is within the permissible limits and do not affect the

performance of the members, a detailed study on the deformation assumes significance.

A quantity called strain defines the deformation of the members and structures in a better

way than the deformation itself and is an indication on the state of the material.

Figure 1.12

Consider a rod of uniform cross section with initial length as shown in figure 1.12.

Application of a tensile load P at one end of the rod results in elongation of the rod by

0

L

δ

.

After elongation, the length of the rod is L. As the cross section of the rod is uniform, it is

appropriate to assume that the elongation is uniform throughout the volume of the rod. If

the tensile load is replaced by a compressive load, then the deformation of the rod will be a

contraction. The deformation per unit length of the rod along its axis is defined as the

normal strain. It is denoted by

ε

0

L L

L L

−

δ

ε = =

1.9

Though the strain is a dimensionless quantity, units are often given in mm/mm, µm/m.

Strength of Materials Prof. M. S. Sivakumar

Indian Institute of Technology Madras

Example 3:

A circular hollow tube made of steel is used to support a compressive load of 500kN. The

inner and outer diameters of the tube are 90mm and 130mm respectively and its length is

1000mm. Due to compressive load, the contraction of the rod is 0.5mm. Determine the

compressive stress and strain in the post.

Solution

Force,

3

P 500 10 N(compressive)= − ×

Area of the tube,

( ) ( )

2 2

3 2

A 0.13 0.09 = 6.912 10 m

4

−

π

⎡ ⎤

= − ×

⎢ ⎥

⎣ ⎦

Stress,

3

3

P 500 10

72.3MPa (compressive)

A

6.912 10

−

− ×

σ = = = −

×

Strain,

4

0

0.5

5 10 (compressive)

L 1000

−

δ −

ε = = = − ×

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Strength of Materials Prof. M. S. Sivakumar

Indian Institute of Technology Madras

1.6 Mechanical properties of materials

A tensile test is generally conducted on a standard specimen to obtain the relationship

between the stress and the strain which is an important characteristic of the material.

In the test, the uniaxial load is applied to the specimen and increased gradually. The

corresponding deformations are recorded throughout the loading.

Stress-strain diagrams of materials vary widely depending upon whether the material is

ductile or brittle in nature.

If the material undergoes a large deformation before failure, it is referred to as ductile

material or else brittle material.

Figure 1.13

In figure 1.13, the stress-strain diagram of a structural steel, which is a ductile material, is

given.

Initial part of the loading indicates a linear relationship between stress and strain, and the

deformation is completely recoverable in this region for both ductile and brittle materials.

This linear relationship, i.e., stress is directly proportional to strain, is popularly known as

Hooke's law.

E

σ

= ε

1.10

The co-efficient E is called the modulus of elasticity or Young's modulus.

Most of the engineering structures are designed to function within their linear elastic region

only.

Strength of Materials Prof. M. S. Sivakumar

Indian Institute of Technology Madras

After the stress reaches a critical value, the deformation becomes irrecoverable. The

corresponding stress is called the yield stress or yield strength of the material beyond

which the material is said to start yielding.

In some of the ductile materials like low carbon steels, as the material reaches the yield

strength it starts yielding continuously even though there is no increment in external

load/stress. This flat curve in stress strain diagram is referred as perfectly plastic region.

The load required to yield the material beyond its yield strength increases appreciably and

this is referred to strain hardening of the material.

In other ductile materials like aluminum alloys, the strain hardening occurs immediately

after the linear elastic region without perfectly elastic region.

After the stress in the specimen reaches a maximum value, called ultimate strength, upon

further stretching, the diameter of the specimen starts decreasing fast due to local

instability and this phenomenon is called necking.

The load required for further elongation of the material in the necking region decreases

with decrease in diameter and the stress value at which the material fails is called the

breaking strength.

In case of brittle materials like cast iron and concrete, the material experiences smaller

deformation before rupture and there is no necking.

Figure 1.14

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Strength of Materials Prof. M. S. Sivakumar

Indian Institute of Technology Madras

1.7 True stress and true strain

In drawing the stress-strain diagram as shown in figure 1.13, the stress was calculated by

dividing the load P by the initial cross section of the specimen.

But it is clear that as the specimen elongates its diameter decreases and the decrease in

cross section is apparent during necking phase.

Hence, the actual stress which is obtained by dividing the load by the actual cross

sectional area in the deformed specimen is different from that of the engineering stress

that is obtained using undeformed cross sectional area as in equation 1.1

True stress or actual stress,

act

act

P

A

σ =

1.11

Though the difference between the true stress and the engineering stress is negligible for

smaller loads, the former is always higher than the latter for larger loads.

Similarly, if the initial length of the specimen is used to calculate the strain, it is called

engineering strain as obtained in equation 1.9

But some engineering applications like metal forming process involve large deformations

and they require actual or true strains that are obtained using the successive recorded

lengths to calculate the strain.

True strain

0

L

0

L

dL L

ln

L L

⎛ ⎞

= =

⎜ ⎟

⎝ ⎠

∫

1.12

True strain is also called as actual strain or natural strain and it plays an important role in

theories of viscosity.

The difference in using engineering stress-strain and the true stress-strain is noticeable

after the proportional limit is crossed as shown in figure 1.15.

Strength of Materials Prof. M. S. Sivakumar

Indian Institute of Technology Madras

Figure 1.15

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Strength of Materials Prof. M. S. Sivakumar

Indian Institute of Technology Madras

1.8 Poissons ratio

Figure 1.16

Consider a rod under an axial tensile load P as shown in figure 1.6 such that the material

is within the elastic limit. The normal stress on x plane is

xx

P

A

σ

=

and the associated

longitudinal strain in the x direction can be found out from

xx

x

E

σ

ε =

. As the material

elongates in the x direction due to the load P, it also contracts in the other two mutually

perpendicular directions, i.e., y and z directions.

Hence, despite the absence of normal stresses in y and z directions, strains do exist in

those directions and they are called lateral strains.

The ratio between the lateral strain and the axial/longitudinal strain for a given material is

always a constant within the elastic limit and this constant is referred to as Poisson's ratio.

It is denoted by .

ν

lateral strain

axial strain

ν = −

1.13

Since the axial and lateral strains are opposite in sign, a negative sign is introduced in

equation 1.13 to make

ν

positive.

Using equation 1.13, the lateral strain in the material can be obtained by

xx

y z x

E

σ

ε = ε = −νε = −ν

1.14

Poisson's ratio can be as low as 0.1 for concrete and as high as 0.5 for rubber.

In general, it varies from 0.25 to 0.35 and for steel it is about 0.3.

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Strength of Materials Prof. M. S. Sivakumar

Indian Institute of Technology Madras

1.9 Elasticity and Plasticity

If the strain disappears completely after removal of the load, then the material is said to be

in elastic region.

The stress-strain relationship in elastic region need not be linear and can be non-linear as

in rubber like materials.

Figure 1.17

The maximum stress value below which the strain is fully recoverable is called the elastic

limit. It is represented by point A in figure 1.17.

When the stress in the material exceeds the elastic limit, the material enters into plastic

phase where the strain can no longer be completely removed.

To ascertain that the material has reached the plastic region, after each load increment, it

is unloaded and checked for residual strain.

Presence of residual strain is the indication that the material has entered into plastic

phase.

If the material has crossed elastic limit, during unloading it follows a path that is parallel to

the initial elastic loading path with the same proportionality constant E.

The strain present in the material after unloading is called the residual strain or plastic

strain and the strain disappears during unloading is termed as recoverable or elastic strain.

They are represented by OC and CD, respectively in figure.1.17.

If the material is reloaded from point C, it will follow the previous unloading path and line

CB becomes its new elastic region with elastic limit defined by point B.

Strength of Materials Prof. M. S. Sivakumar

Indian Institute of Technology Madras

Though the new elastic region CB resembles that of the initial elastic region OA, the

internal structure of the material in the new state has changed.

The change in the microstructure of the material is clear from the fact that the ductility of

the material has come down due to strain hardening.

When the material is reloaded, it follows the same path as that of a virgin material and fails

on reaching the ultimate strength which remains unaltered due to the intermediate loading

and unloading process.

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Strength of Materials Prof. M. S. Sivakumar

Indian Institute of Technology Madras

1.10 Creep and fatigue

In the preceding section, it was discussed that the plastic deformation of a material

increases with increasing load once the stress in the material exceeds the elastic limit.

However, the materials undergo additional plastic deformation with time even though the

load on the material is unaltered.

Consider a bar under a constant axial tensile load as shown in figure 1.18.

Figure 1.18

As soon as the material is loaded beyond its elastic limit, it undergoes an instant plastic

deformation at time t = 0.

0

ε

Though the material is not brought under additional loads, it experiences further plastic

deformation with time as shown in the graph in figure 1.18.

This phenomenon is called creep.

Creep at high temperature is of more concern and it plays an important role in the design

of engines, turbines, furnaces, etc.

However materials like concrete, steel and wood experience creep slightly even at normal

room temperature that is negligible.

Analogous to creep, the load required to keep the material under constant strain

decreases with time and this phenomenon is referred to as stress relaxation.

It was concluded in section 1.9 that the specimen will not fail when the stress in the

material is with in the elastic limit.

Strength of Materials Prof. M. S. Sivakumar

Indian Institute of Technology Madras

This holds true only for static loading conditions and if the applied load fluctuates or

reverses then the material will fail far below its yield strength.

This phenomenon is known as fatigue.

Designs involving fluctuating loads like traffic in bridges, and reversing loads like

automobile axles require fatigue analysis.

Fatigue failure is initiated by a minute crack that develops at a high stress point which may

be an imperfection in the material or a surface scratch.

The crack enlarges and propagates through the material due to successive loadings until

the material fails as the undamaged portion of the material is insufficient to withstand the

load.

Hence, a polished surface shaft can take more number of cycles than a shaft with rough or

corroded surface.

The number of cycles that can be taken up by a material before it fractures can be found

out by conducting experiments on material specimens.

The obtained results are plotted as

n

σ

−

curves as given in figure 1.19, which indicates the

number of cycles that can be safely completed by the material under a given maximum

stress.

Figure 1.19

It is learnt from the graph that the number of cycles to failure increases with decrease in

magnitude of stress.

Strength of Materials Prof. M. S. Sivakumar

Indian Institute of Technology Madras

For steels, if the magnitude of stress is reduced to a particular value, it can undergo an

infinitely large number of cycles without fatigue failure and the corresponding stress is

known as endurance limit or fatigue limit.

On the other hand, for non-ferrous metals like aluminum alloys there is no endurance limit,

and hence, the maximum stress decreases continuously with increase in number of cycles.

In such cases, the fatigue limit of the material is taken as the stress value that will allow an

arbitrarily taken number of cycles, say cycles.

8

10

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Strength of Materials Prof. M. S. Sivakumar

Indian Institute of Technology Madras

1.11 Deformation in axially loaded members

Consider the rod of uniform cross section under tensile load P along its axis as shown in

figure 1.12.

Let that the initial length of the rod be L and the deflection due to load be

δ

. Using

equations 1.9 and 1.10,

P

L E A

PL

AE

δ σ

= ε = =

δ =

E

1.15

Equation 1.15 is obtained under the assumption that the material is homogeneous and has

a uniform cross section.

Now, consider another rod of varying cross section with the same axial load P as shown in

figure 1.20.

Figure 1.20

Let us take an infinitesimal element of length dx in the rod that undergoes a deflection

d due to load P. The strain in the element is

δ

d

and d = dx

dx

δ

ε

= δ

ε

The deflection of total length of the rod can be obtained by integrating above equation,

dx

δ

= ε

∫

L

0

Pdx

EA(x)

δ =

∫

1.16

Strength of Materials Prof. M. S. Sivakumar

Indian Institute of Technology Madras

As the cross sectional area of the rod keeps varying, it is expressed as a function of its

length.

If the load is also varying along the length like the weight of the material, it should also be

expressed as a function of distance, i.e., P(x) in equation 1.16.

Also, if the structure consists of several components of different materials, then the

deflection of each component is determined and summed up to get the total deflection of

the structure.

When the cross section of the components and the axial loads on them are not varying

along length, the total deflection of the structure can be determined easily by,

n

i i

i i

i 1

P L

A E

=

δ =

∑

1.17

Example 4:

Figure 1.21

Consider a rod ABC with aluminum part AB and steel part BC having diameters 25mm and

50 mm respectively as shown figure 1.21. Determine the deflections of points A and B.

Strength of Materials Prof. M. S. Sivakumar

Indian Institute of Technology Madras

Solution:

Deflection of part AB =

3

AB AB

2 9

AB AB

P L

10 10 400

A E

(0.025)/4 70 10

× ×

= −

π× × ×

0.1164 mm= −

Deflection of part BC =

3

BC BC

2 9

BC BC

P L 35 10 500

A E

(0.05)/4 200 10

× ×

= −

π× × ×

0.0446 mm= −

Deflection point of B

0.0446 mm= −

Deflection point of A

(

)

( 0.1164) 0.0446= − + −

0.161mm= −

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Strength of Materials Prof. M. S. Sivakumar

Indian Institute of Technology Madras

1.12 Statically indeterminate problems.

Members for which reaction forces and internal forces can be found out from static

equilibrium equations alone are called statically determinate members or structures.

Problems requiring deformation equations in addition to static equilibrium equations to

solve for unknown forces are called statically indeterminate problems.

Figure 1.22

The reaction force at the support for the bar ABC in figure 1.22 can be determined

considering equilibrium equation in the vertical direction.

y

F 0; R - P = 0=

∑

Now, consider the right side bar MNO in figure 1.22 which is rigidly fixed at both the ends.

From static equilibrium, we get only one equation with two unknown reaction forces R

1

and

R

2.

1 2

- P + R + R = 0

1.18

Hence, this equilibrium equation should be supplemented with a deflection equation which

was discussed in the preceding section to solve for unknowns.

If the bar MNO is separated from its supports and applied the forces, then

these forces cause the bar to undergo a deflection

1 2

R,R and P

MO

δ

that must be equal to zero.

Strength of Materials Prof. M. S. Sivakumar

Indian Institute of Technology Madras

MO MN NO

0δ = ⇒δ + δ =

0

1.19

MN NO

and δ δ

are the deflections of parts MN and NO respectively in the bar MNO.

Individually these deflections are not zero, but their sum must make it to be zero.

Equation 1.19 is called compatibility equation, which insists that the change in length of the

bar must be compatible with the boundary conditions.

Deflection of parts MN and NO due to load P can be obtained by assuming that the

material is within the elastic limit,

1 1 2 2

MN NO

1 2

R l R l

and

A E A E

δ = δ =

.

Substituting these deflections in equation 1.19,

1 1 2 2

1 2

R l R l

- 0

A E A E

=

1.20

Combining equations 1.18 and 1.20, one can get,

1 2

1

1 2 2 1

2 1

2

1 2 2 1

PA l

R

l A l A

PA l

R

l A l A

=

+

=

+

1.21

From these reaction forces, the stresses acting on any section in the bar can be easily

determined.

Example 5:

Figure 1.23

Strength of Materials Prof. M. S. Sivakumar

Indian Institute of Technology Madras

A rectangular column of sides

0.4m 0.35m

×

, made of concrete, is used to support a

compressive load of 1.5MN. Four steel rods of each 24mm diameter are passing through

the concrete as shown in figure 1.23. If the length of the column is 3m, determine the

normal stress in the steel and the concrete. Take

steel concrete

E 200 GPa and E 29 GPa.

=

=

Solution:

P

s

= Load on each steel rod

P

c

= Load on concrete

From equilibrium equation,

c s

3

c s

P 4P P

P 4P 1.5 10..........(a)

+ =

+ = ×

Deflection in steel rod and concrete are the same.

concrete steel

δ = δ

( )

c s

9

2 9

c s

P 3 P 3

0.4 (0.35) 29 10

(0.024) 200 10

4

P 44.87P................(b)

× ×

=

π

× × ×

× × ×

=

Combining equations (a) and (b),

s

c

P 30.7kN

P 1378kN

=

=

Normal stress on concrete=

6

c

c

P

1.378 10

9.84MPa

A (0.4)(0.35)

×

= =

Normal stress on steel=

3

s

2

s

P

30.7 10

67.86MPa

A

(0.024)

4

×

= =

π

Top

Strength of Materials Prof. M. S. Sivakumar

Indian Institute of Technology Madras

1.13 Thermal effect

When a material undergoes a change in temperature, it either elongates or contracts

depending upon whether heat is added to or removed from the material.

If the elongation or contraction is not restricted, then the material does not experience any

stress despite the fact that it undergoes a strain.

The strain due to temperature change is called thermal strain and is expressed as

T

( T)

ε

= α ∆

1.22

where

α

is a material property known as coefficient of thermal expansion and

∆

T indicates

the change in temperature.

Since strain is a dimensionless quantity and

∆

T is expressed in K or

0

C,

α

has a unit that is

reciprocal of K or

0

C.

The free expansion or contraction of materials, when restrained induces stress in the

material and it is referred to as thermal stress.

Thermal stress produces the same effect in the material similar to that of mechanical

stress and it can be determined as follows.

Figure 1.24

Consider a rod AB of length L which is fixed at both ends as shown in figure 1.24.

Let the temperature of the rod be raised by

∆

T and as the expansion is restricted, the

material develops a compressive stress.

Strength of Materials Prof. M. S. Sivakumar

Indian Institute of Technology Madras

In this problem, static equilibrium equations alone are not sufficient to solve for unknowns

and hence is called statically indeterminate problem.

To determine the stress due to

∆

T, assume that the support at the end B is removed and

the material is allowed to expand freely.

Increase in the length of the rod

T

δ

due to free expansion can be found out using equation

1.22

T T

L ( T)δ = ε = α ∆

L

1.23

Now, apply a compressive load P at the end B to bring it back to its initial position and the

deflection due to mechanical load from equation 1.15,

T

PL

AE

δ =

1.24

As the magnitude of and

δ

are equal and their signs differ,

T

δ

T

PL

( T)L

AE

δ = −δ

α ∆ = −

T

P

Thermal stress, ( T)E

A

σ = = −α ∆

1.25

Minus sign in the equation indicates a compressive stress in the material and with

decrease in temperature, the stress developed is tensile stress as

∆

T becomes negative.

It is to be noted that the equation 1.25 was obtained on the assumption that the material is

homogeneous and the area of the cross section is uniform.

Thermoplastic analysis assumes significance for structures and components that are

experiencing high temperature variations.

Strength of Materials Prof. M. S. Sivakumar

Indian Institute of Technology Madras

Example 6:

Figure 1.25

A rod consists of two parts that are made of steel and aluminum as shown in figure 1.25.

The elastic modulus and coefficient of thermal expansion for steel are 200GPa and 11.7 x

10

-6

per

0

C respectively and for aluminum 70GPa and 21.6 x 10

-6

per

0

C respectively. If the

temperature of the rod is raised by 50

0

C, determine the forces and stresses acting on the

rod.

Solution:

Deflection of the rod under free expansion,

( ) (

T

6 6

( T)L

11.7 10 50 500 21.6 10 50 750

1.1025 mm

− −

δ = α ∆

= × × × + × × ×

=

)

Restrained deflection of rod = 1.1025 - 0.4 = 0.7025 mm

Let the force required to make their elongation vanish be R which is the reaction force at

the ends.

Strength of Materials Prof. M. S. Sivakumar

Indian Institute of Technology Madras

[ ]

[ ]

steel Al

2

3 2

2

3 2

3 9 3

RL RL

AE AE

Area of steel rod = 0.05 1.9635 10 m

4

Area of aluminium rod = 0.03 0.7069 10 m

4

500 750

0.7025 R

1.9635 10 200 10 0.7069 10 70 10

Compressive force on the r

−

−

− −

⎛ ⎞ ⎛ ⎞

−δ = +

⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

π

= ×

π

= ×

⎡ ⎤

− = +

⎢ ⎥

× × × × × ×⎣ ⎦

9

3

3

3

3

od, R = 42.076 kN

P 42.76 10

Compressive stress on steel, = 21.8MPa

A

1.9635 10

P 42.76 10

Compressive stress on steel, = 60.5MPa

A

0.7069 10

−

−

−

− ×

σ = = −

×

− ×

σ = = −

×

Top

Strength of Materials Prof. M. S. Sivakumar

Indian Institute of Technology Madras

1.14. Design considerations

:

A good design of a structural element or machine component should ensure that the

developed product will function safely and economically during its estimated life time.

The stress developed in the material should always be less than the maximum stress it

can withstand which is known as ultimate strength as discussed in section 1.6.

During normal operating conditions, the stress experienced by the material is referred to as

working stress or allowable stress or design stress.

The ratio of ultimate strength to allowable stress is defined as factor of safety.

Ultimatestress

Factor of safety

Allowablestress

=

1.26

Factor of safety can also be expressed in terms of load as,

Ultimateload

Factor of safety

Allowableload

=

1.27

Equations 1.26 and 1.27 are identical when a linear relationship exists between the load

and the stress.

This is not true for many materials and equation 1.26 is widely used in design analysis.

Factor of safety take care of the uncertainties in predicting the exact loadings, variation in

material properties, environmental effects and the accuracy of methods of analysis.

If the factor of safety is less, then the risk of failure is more and on the other hand, when

the factor of safety is very high the structure becomes unacceptable or uncompetitive.

Hence, depending upon the applications the factor of safety varies. It is common to see

that the factor of safety is taken between 2 and 3.

Stresses developed in the material when subjected to loads can be considered to be

uniform at sections located far away from the point of application of loads.

This observation is called Saint Venant’s principle and was discussed in section 1.3.

But, when the element has holes, grooves, notches, key ways, threads and other

abrupt changes in geometry, the stress on those cross-sections will not be uniform.

Strength of Materials Prof. M. S. Sivakumar

Indian Institute of Technology Madras

These discontinuities in geometry cause high stresses concentrations in small regions of

the material and are called stress raisers.

Experimentally it was found that the stress concentrations are independent of the material

size and its properties, and they depend only on the geometric parameters.

Figure 1.26

Consider a rectangular flat plate with a circular hole as shown in figure 1.26.

The stress distribution on the section passing through the centre of the hole indicates that

the maximum stress occurs at the ends of the holes and it is much higher than the average

stress.

Since the designer, in general, is more interested in knowing the maximum stress rather

than the actual stress distribution, a simple relationship between the in

terms of geometric parameters will be of practical importance.

max ave

andσ σ

Many experiments were conducted on samples with various discontinuities and the

relationship between the stress concentration factor and the geometrical parameters are

established, where

Stress concentration factor,

max

ave

K

σ

=

σ

1.28

Strength of Materials Prof. M. S. Sivakumar

Indian Institute of Technology Madras

Hence, simply by calculating the average stress,

ave

P

A

σ

=

, in the critical section of a

discontinuity, can be easily found and by multiplying

max

σ

ave

σ

with K.

The variation of K in terms of r/d for the rectangular plate with a circular hole is given in

figure 1.26.

It is to be noted that the expression in equation 1.28 can be used as long as is within

the proportional limit of the material.

max

σ

Example 7:

Figure 1.27

A rectangular link AB made of steel is used to support a load W through a rod CD as

shown in figure 1.27. If the link AB is 30mm wide, determine its thickness for a factor of

safety 2.5. The ultimate strength of steel may be assumed to be 450 MPa.

Solution:

Drawing free body diagram of the link and the rod,

Taking moment about C,

Strength of Materials Prof. M. S. Sivakumar

Indian Institute of Technology Madras

0

BA

BA

BA

6

a

3

6

25 550 F sin60 500 0

F 31.75 kN

Tensionalonglink AB, F 31.75 kN.

Ultimatestress

F.O.S

allowablestress

450 10

Allowablestressinlink AB,180MPa

2.5

Tensileforce

Stressinlink AB,

Area

31.75 10

180 10

0.03

× − × =

=

=

=

×

σ = =

σ =

×

× =

t

Thicknessof link AB,t 5.88

t 6mm

×

=

≈

Top

Strength of Materials Prof. M. S. Sivakumar

Indian Institute of Technology Madras

1.15. Strain energy:

Strain energy is an important concept in mechanics and is used to study the response of

materials and structures under static and dynamic loads.

Within the elastic limit, the work done by the external forces on a material is stored as

deformation or strain that is recoverable.

On removal of load, the deformation or strain disappears and the stored energy is

released. This recoverable energy stored in the material in the form of strain is called

elastic strain energy.

Figure 1.28

Consider a rod of uniform cross section with length L as shown in figure 1.28.

An axial tensile load P is applied on the material gradually from zero to maximum

magnitude and the corresponding maximum deformation is δ.

Area under the load-displacement curve shown in figure 1.28 indicates the work done on

the material by the external load that is stored as strain energy in the material.

Let dW be the work done by the load P due to increment in deflection dδ. The

corresponding increase in strain energy is dU.

When the material is within the elastic limit, the work done due to dδ,

dW dU Pd= =

δ

The total work done or total elastic strain energy of the material,

0

W U Pd

δ

=

=

δ

∫

1.29

Strength of Materials Prof. M. S. Sivakumar

Indian Institute of Technology Madras

Equation 1.29 holds for both linear elastic and non-linear elastic materials.

If the material is linear elastic, then the load-displacement diagram will become as shown

in figure 1.29.

Figure 1.29

The elastic strain energy stored in the material is determined from the area of triangle

OAB.

1 1

1

U P

2

=

δ

1.30

1

1

P L

where

AE

δ =

.

Since the load-displacement curve is a straight line here, the load can be expressed in

terms of stiffness and deflection as

1

P

1

P k

1

=

δ

. Then equation 1.30 turns out to be,

2

1

1

U k

2

=

δ

1.31

Work done and strain energy are expressed in N-m or joules (J).

Strain energy defined in equation 1.29 depends on the material dimensions.

In order to eliminate the material dimensions from the strain energy equation, strain energy

density is often used.

Strain energy stored per unit volume of the material is referred to as strain energy density.

Dividing equation 1.29 by volume,

Strength of Materials Prof. M. S. Sivakumar

Indian Institute of Technology Madras

Strain energy density,

0

u

d

ε

=

σ ε

∫

1.32

Equation 1.32 indicates the expression of strain energy in terms of stress and strain, which

are more convenient quantities to use rather than load and displacement.

Figure 1.30

Area under the stress strain curve indicates the strain energy density of the material.

For linear elastic materials within proportional limit, equation 1.32 gets simplified as,

Strain energy density,

1 1

1

u

2

=

σ ε

1.33

Using Hook’s law,

1

1

E

σ

ε =

, strain energy density is expressed in terms of stress,

2

1

u

2E

σ

=

1.34

When the stress in the material reaches the yield stress

y

σ

, the strain energy density

attains its maximum value and is called the modulus of resilience.

Modulus of resilience,

2

Y

R

u

2E

σ

=

1.35

Strength of Materials Prof. M. S. Sivakumar

Indian Institute of Technology Madras

Modulus of resilience is a measure of energy that can be absorbed by the material due to

impact loading without undergoing any plastic deformation.

Figure 1.31

If the material exceeds the elastic limit during loading, all the work done is not stored in the

material as strain energy.

This is due to the fact that part of the energy is spent on deforming the material

permanently and that energy is dissipated out as heat.

The area under the entire stress strain diagram is called modulus of toughness, which is a

measure of energy that can be absorbed by the material due to impact loading before it

fractures.

Hence, materials with higher modulus of toughness are used to make components and

structures that will be exposed to sudden and impact loads.

Example 8:

Figure 1.32

Strength of Materials Prof. M. S. Sivakumar

Indian Institute of Technology Madras

A 25 kN load is applied gradually on a steel rod ABC as shown in figure 1.32. Taking

E=200 GPa, determine the strain energy stored in the entire rod and the strain energy

density in parts AB and BC. If the yield strength of the material is σ

y

=320MPa, find the

maximum energy that can be absorbed by the rod without undergoing permanent

deformation.

Solution:

( )

2

AB

AB

2

3

AB

9

2

3

Strain energy density in part AB, u

2E

1 25 10

u

2 200 10

0.024

4

= 7.63 kJ/m

σ

=

⎡

⎤

⎢

⎥

×

=

⎢

⎥

π

× ×

⎢

⎥

⎣

⎦

( )

2

BC

BC

2

3

BC

9

2

3

Strain energy density in part BC, u

2E

1 25 10

u

2 200 10

0.016

4

= 38.65 kJ/m

σ

=

⎡

⎤

⎢

⎥

×

=

⎢

⎥

π

× ×

⎢

⎥

⎣

⎦

Strain energy in the entire rod,

( )

( )

AB AB BC BC

2 2

3 3

U u V u V

7.63 10 0.024 1 38.65 10 0.016 0.8

4 4

U 9.67J

= +

Π Π

⎡ ⎤ ⎡

= × × × + × × ×

⎢ ⎥ ⎢

⎣ ⎦ ⎣

=

⎤

⎥

⎦

The load that will produce yield stress in the material,

( )

2

6

y BC

P A 320 10 0.016

4

P 64.3kN

Π

= σ = × ×

=

Maximum energy that can be stored in the rod

,

Strength of Materials Prof. M. S. Sivakumar

Indian Institute of Technology Madras

( )

( )

( )

22

AB BC

AB BC

2

BCAB

AB BC

2

3

9

2 2

1 P P

U V V

2E A A

LLP

2E A A

64.3 10

1 0.8

2 200 10

0.024 0.016

4 4

63.97J

⎡ ⎤

⎛ ⎞

⎛ ⎞

⎢ ⎥

= × + ×

⎜ ⎟⎜ ⎟

⎢ ⎥

⎝ ⎠

⎝ ⎠

⎣ ⎦

⎡ ⎤

= +

⎢ ⎥

⎣ ⎦

⎡ ⎤

×

⎢ ⎥

= +

⎢ ⎥

Π Π

× ×

⎢ ⎥

⎣ ⎦

=

Top

Strength of Materials Prof. M. S. Sivakumar

Indian Institute of Technology Madras

1.16 Impact loading

A static loading is applied very slowly so that the external load and the internal force are

always in equilibrium. Hence, the vibrational and dynamic effects are negligible in static

loading.

Dynamic loading may take many forms like fluctuating loads where the loads are varying

with time and impact loads where the loads are applied suddenly and may be removed

immediately or later.

Collision of two bodies and objects freely falling onto a structure are some of the examples

of impact loading.

Consider a collar of mass M at a height h from the flange that is rigidly fixed at the end of a

bar as shown in figure 1.33.

As the collar freely falls onto the flange, the bar begins to elongate causing axial stresses

and strain within the bar

.

Figure 1.33

After the flange reaching its maximum position during downward motion, it moves up due

to shortening of the bar.

The bar vibrates in the axial direction with the collar and the flange till the vibration dies out

completely due to damping effects.

To simplify the complex impact loading analysis, the following assumptions are made.

Strength of Materials Prof. M. S. Sivakumar

Indian Institute of Technology Madras

The kinetic energy of the collar at the time of striking is completely transformed into strain

energy and stored in the bar.

But in practice, not all the kinetic energy is stored in the material as some of the energy is

dissipated out as heat and waves.

Hence, this assumption is conservative in the sense that the stress and deflection

predicted by this way are higher than the actual values.

The second assumption is that after striking the flange, the collar and the flange move

downward together without any bouncing.

This assumption is reasonable provided the weight of the collar is much larger than that of

the bar.

The third assumption is that the stresses in the bar remain within linear elastic range and

the stress distribution is uniform within the bar.

But, in reality, the stress distribution is not uniform since the stress waves generated due

to impact loading travel through the bar.

Using the principle of conservation of energy, the kinetic energy of the collar is equated to

the strain energy of the bar.

Assuming the height of fall h is much larger than the deformation of rod, and using

equation 1.34,

2

2

max

1

2 2

V

Mv

E

σ

=

1.36

where v is the velocity of the collar at strike (

2ghv =

) and V is the volume of the material.

The maximum stress in the bar due to the impact load of mass M,

2

max

M

v E

V

σ =

1.37

From above equation, it becomes clear that by increasing the volume of material, the effect

of impact loading can be minimized.

Expressing strain energy in terms of deflection in equation 1.36,

2

2

max

1

2 2

EA

Mv

L

δ

=

Strength of Materials Prof. M. S. Sivakumar

Indian Institute of Technology Madras

2

max

M

v L

E

A

δ =

1.38

If the load of the collar is applied gradually on the bar i.e., under static loading, the static

deflection

δ

st

will be,

st

MgL

EA

δ =

Substituting this in equation 1.38, relationship between the static deflection

δ

st

and the

impact deflection

δ

max

is obtained.

max st

2h

δ

= δ

1.39

To represent the magnification of deflection due to impact load compared to that of static

deflection for the same load, impact factor is used.

max

st

Impact factor

δ

=

δ

1.40

Alternately, the impact factor can be obtained from the ratio

max

st

σ

σ

.

The relationship between the stress

st

σ

developed in the bar due to static loading and the

impact loading stress is determined as follows.

max

σ

st

st st

max

max max

max st

E

= E =

L

E

= E =

L

E

= 2h

L

δ

σ ε

δ

σ ε

σ δ

st

max

2hE

=

L

σ

σ

1.41

Now, the effect of suddenly applied loads on materials or structures that forms a special

case of impact loading is discussed.

Strength of Materials Prof. M. S. Sivakumar

Indian Institute of Technology Madras

In figure 1.33, if the collar is brought into contact on the top of the flange and released

immediately, it is referred to as suddenly applied load.

The maximum stress produced in the bar due to suddenly applied load can be determined

by replacing h by

δ

max

in equation 1.41

max st

= 2

σ

σ

1.42

Hence, the stress developed in a material due to suddenly applied load is twice as large as

that of gradually applied load.

Example 9:

Figure 1.34

A 50 kg collar is sliding on a cable as shown in figure 1.34 from a height h = 1m. Its free

fall is restrained by a stopper at the end of the cable. The effective cross-sectional area

and the elastic modulus of the cable are taken to be 60 mm

2

and 150GPa respectively. If

the maximum allowable stress in the cable due to impact load is 450MPa, calculate the

minimum permissible length for the cable and the corresponding maximum deflection. Also

find the impact factor.

Solution:

Maximum stress due to impact load,

2

max

M

=

v E

V

σ

Strength of Materials Prof. M. S. Sivakumar

Indian Institute of Technology Madras

Velocity, v

=

2gh = 2 9.81 1 = 4.43 m/s× ×

( )

( )

2

9

6

6

50 4.43 150 10

450 10 =

60 10 L

−

× × ×

×

× ×

Minimum permissible length for the cable, L = 12.1 m

Static deflection,

st

MgL

EA

δ =

9 6

50 9.81 12.1

150 10 60 10

−

×

×

=

×

× ×

= 0.656 mm

Maximum deflection,

max st

2h

δ

= δ

2 1000 0.656

36mm

= × ×

=

Impact factor,

max

st

36

0.656

δ

= =

δ

= 55

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